Question

At constant volume, 1.0 kJ of heat is transferred to 2 moles of a monoatomic ideal gas at 1 atm and 298 K. The final temperature of the ideal gas is P K. [Given: R=8.314 JKmol] (Nearest Integer)

Answer 1

338 K

Answer 2

At constant volume, the heat added increases the internal energy:

$$Q = nC_v\Delta T \quad \text{with} \quad C_v = \frac{3}{2}R \quad \text{(for a monoatomic ideal gas)}$$

Solve for $\Delta T$:

$$\Delta T = \frac{Q}{n\,\frac{3}{2}R} = \frac{1000\,\text{J}}{2 \times \frac{3}{2}\times8.314\,\text{J/(K·mol)}} = \frac{1000}{3 \times 8.314} \approx \frac{1000}{24.942} \approx 40.08\,\text{K}$$

Final temperature:

$$T_{\text{final}} = T_{\text{initial}} + \Delta T = 298\,\text{K} + 40.08\,\text{K} \approx 338\,\text{K}$$